Applying the divided difference operator












1












$begingroup$


This question is about divided difference operators.



How do I perform $partial _2$ or $partial_3$ on $x_1^2x_2$?



$partial_i$ is defined as $frac{p-r_i.p}{x_i-x_{i+1}}$, where $r_i$ is the reflection operator; i.e. $r_i.p(x_1,x_2,dots,x_i,x_{i+1},dots,x_n)=p(x_1,x_2,dots,x_{i+1},x_i,dots,x_n)$.



Do I just assume that $x_1^2x_2=x_1^2x_2x_3^0x_4^0$?



Edit: I have another question. The calculation of Schubert polynomials uss the following formula: $S_{pi}=partial_{pi^{-1}w_0^n}S_{w_0^n}$. What does $partial_{pi^{-1}w_0^n}$ mean? Do I just concatenate $pi^{-1}$ and $w_0^n$ as words? Like if $w_0^3=321$ and $pi=(12)$, we get $partial_{pi^{-1}w_0^n}=partial_{1321}$?






combinatorics representation-theory schubert-calculus







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  • $begingroup$
    See my answer to this mathoverflow question: mathoverflow.net/questions/216242/free-kx-1-dots-x-ns-n-module/…
    $endgroup$
    – David Hill
    May 4 '16 at 14:30
















1












$begingroup$


This question is about divided difference operators.



How do I perform $partial _2$ or $partial_3$ on $x_1^2x_2$?



$partial_i$ is defined as $frac{p-r_i.p}{x_i-x_{i+1}}$, where $r_i$ is the reflection operator; i.e. $r_i.p(x_1,x_2,dots,x_i,x_{i+1},dots,x_n)=p(x_1,x_2,dots,x_{i+1},x_i,dots,x_n)$.



Do I just assume that $x_1^2x_2=x_1^2x_2x_3^0x_4^0$?



Edit: I have another question. The calculation of Schubert polynomials uss the following formula: $S_{pi}=partial_{pi^{-1}w_0^n}S_{w_0^n}$. What does $partial_{pi^{-1}w_0^n}$ mean? Do I just concatenate $pi^{-1}$ and $w_0^n$ as words? Like if $w_0^3=321$ and $pi=(12)$, we get $partial_{pi^{-1}w_0^n}=partial_{1321}$?






combinatorics representation-theory schubert-calculus







share|cite|improve this question











$endgroup$












  • $begingroup$
    See my answer to this mathoverflow question: mathoverflow.net/questions/216242/free-kx-1-dots-x-ns-n-module/…
    $endgroup$
    – David Hill
    May 4 '16 at 14:30














1












1








1


1



$begingroup$


This question is about divided difference operators.



How do I perform $partial _2$ or $partial_3$ on $x_1^2x_2$?



$partial_i$ is defined as $frac{p-r_i.p}{x_i-x_{i+1}}$, where $r_i$ is the reflection operator; i.e. $r_i.p(x_1,x_2,dots,x_i,x_{i+1},dots,x_n)=p(x_1,x_2,dots,x_{i+1},x_i,dots,x_n)$.



Do I just assume that $x_1^2x_2=x_1^2x_2x_3^0x_4^0$?



Edit: I have another question. The calculation of Schubert polynomials uss the following formula: $S_{pi}=partial_{pi^{-1}w_0^n}S_{w_0^n}$. What does $partial_{pi^{-1}w_0^n}$ mean? Do I just concatenate $pi^{-1}$ and $w_0^n$ as words? Like if $w_0^3=321$ and $pi=(12)$, we get $partial_{pi^{-1}w_0^n}=partial_{1321}$?






combinatorics representation-theory schubert-calculus







share|cite|improve this question











$endgroup$




This question is about divided difference operators.



How do I perform $partial _2$ or $partial_3$ on $x_1^2x_2$?



$partial_i$ is defined as $frac{p-r_i.p}{x_i-x_{i+1}}$, where $r_i$ is the reflection operator; i.e. $r_i.p(x_1,x_2,dots,x_i,x_{i+1},dots,x_n)=p(x_1,x_2,dots,x_{i+1},x_i,dots,x_n)$.



Do I just assume that $x_1^2x_2=x_1^2x_2x_3^0x_4^0$?



Edit: I have another question. The calculation of Schubert polynomials uss the following formula: $S_{pi}=partial_{pi^{-1}w_0^n}S_{w_0^n}$. What does $partial_{pi^{-1}w_0^n}$ mean? Do I just concatenate $pi^{-1}$ and $w_0^n$ as words? Like if $w_0^3=321$ and $pi=(12)$, we get $partial_{pi^{-1}w_0^n}=partial_{1321}$?






combinatorics representation-theory schubert-calculus




combinatorics representation-theory schubert-calculus






share|cite|improve this question















share|cite|improve this question













share|cite|improve this question




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edited Jan 5 at 15:50









Matt Samuel

38.8k63769




38.8k63769










asked May 4 '16 at 12:31









fierydemonfierydemon

4,58522158




4,58522158












  • $begingroup$
    See my answer to this mathoverflow question: mathoverflow.net/questions/216242/free-kx-1-dots-x-ns-n-module/…
    $endgroup$
    – David Hill
    May 4 '16 at 14:30


















  • $begingroup$
    See my answer to this mathoverflow question: mathoverflow.net/questions/216242/free-kx-1-dots-x-ns-n-module/…
    $endgroup$
    – David Hill
    May 4 '16 at 14:30
















$begingroup$
See my answer to this mathoverflow question: mathoverflow.net/questions/216242/free-kx-1-dots-x-ns-n-module/…
$endgroup$
– David Hill
May 4 '16 at 14:30




$begingroup$
See my answer to this mathoverflow question: mathoverflow.net/questions/216242/free-kx-1-dots-x-ns-n-module/…
$endgroup$
– David Hill
May 4 '16 at 14:30










1 Answer
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1












$begingroup$

For the first question: Yes, $x_1^2x_2 = x_1^2x_2^1x_3^0x_4^0$, hence
$$partial_2(x_1^2x_2) = frac{x_1^2x_2 - x_1^2x_3}{x_2-x_3} = x_1^2.$$



For the second question: No, it is not just concatenation. To compute $partial_w$ you need to find a reduced word $I = [i_1, i_2, ldots, i_m]$ for $w$: write $w = r_{i_1}r_{i_2} dotsb r_{i_m}$ with $m$ minimal. Then $partial_w = partial_{i_1}partial_{i_2} dotsb partial_{i_m}.$



In your example, with permutations written in second-line notation, $w_0^3 = 321$ and $pi = 213$, hence $pi^{-1}w_0^3 = 312 = r_2r_1$. Therefore $partial_{pi^{-1}w_0^3} = partial_2partial_1$.






share|cite|improve this answer











$endgroup$













  • $begingroup$
    To Ayush: If my answer is not clear enough for you to accept it, please let me know what else should I add. Thank you.
    $endgroup$
    – Catalin Zara
    Nov 30 '16 at 21:06











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$begingroup$

For the first question: Yes, $x_1^2x_2 = x_1^2x_2^1x_3^0x_4^0$, hence
$$partial_2(x_1^2x_2) = frac{x_1^2x_2 - x_1^2x_3}{x_2-x_3} = x_1^2.$$



For the second question: No, it is not just concatenation. To compute $partial_w$ you need to find a reduced word $I = [i_1, i_2, ldots, i_m]$ for $w$: write $w = r_{i_1}r_{i_2} dotsb r_{i_m}$ with $m$ minimal. Then $partial_w = partial_{i_1}partial_{i_2} dotsb partial_{i_m}.$



In your example, with permutations written in second-line notation, $w_0^3 = 321$ and $pi = 213$, hence $pi^{-1}w_0^3 = 312 = r_2r_1$. Therefore $partial_{pi^{-1}w_0^3} = partial_2partial_1$.






share|cite|improve this answer











$endgroup$













  • $begingroup$
    To Ayush: If my answer is not clear enough for you to accept it, please let me know what else should I add. Thank you.
    $endgroup$
    – Catalin Zara
    Nov 30 '16 at 21:06
















1












$begingroup$

For the first question: Yes, $x_1^2x_2 = x_1^2x_2^1x_3^0x_4^0$, hence
$$partial_2(x_1^2x_2) = frac{x_1^2x_2 - x_1^2x_3}{x_2-x_3} = x_1^2.$$



For the second question: No, it is not just concatenation. To compute $partial_w$ you need to find a reduced word $I = [i_1, i_2, ldots, i_m]$ for $w$: write $w = r_{i_1}r_{i_2} dotsb r_{i_m}$ with $m$ minimal. Then $partial_w = partial_{i_1}partial_{i_2} dotsb partial_{i_m}.$



In your example, with permutations written in second-line notation, $w_0^3 = 321$ and $pi = 213$, hence $pi^{-1}w_0^3 = 312 = r_2r_1$. Therefore $partial_{pi^{-1}w_0^3} = partial_2partial_1$.






share|cite|improve this answer











$endgroup$













  • $begingroup$
    To Ayush: If my answer is not clear enough for you to accept it, please let me know what else should I add. Thank you.
    $endgroup$
    – Catalin Zara
    Nov 30 '16 at 21:06














1












1








1





$begingroup$

For the first question: Yes, $x_1^2x_2 = x_1^2x_2^1x_3^0x_4^0$, hence
$$partial_2(x_1^2x_2) = frac{x_1^2x_2 - x_1^2x_3}{x_2-x_3} = x_1^2.$$



For the second question: No, it is not just concatenation. To compute $partial_w$ you need to find a reduced word $I = [i_1, i_2, ldots, i_m]$ for $w$: write $w = r_{i_1}r_{i_2} dotsb r_{i_m}$ with $m$ minimal. Then $partial_w = partial_{i_1}partial_{i_2} dotsb partial_{i_m}.$



In your example, with permutations written in second-line notation, $w_0^3 = 321$ and $pi = 213$, hence $pi^{-1}w_0^3 = 312 = r_2r_1$. Therefore $partial_{pi^{-1}w_0^3} = partial_2partial_1$.






share|cite|improve this answer











$endgroup$



For the first question: Yes, $x_1^2x_2 = x_1^2x_2^1x_3^0x_4^0$, hence
$$partial_2(x_1^2x_2) = frac{x_1^2x_2 - x_1^2x_3}{x_2-x_3} = x_1^2.$$



For the second question: No, it is not just concatenation. To compute $partial_w$ you need to find a reduced word $I = [i_1, i_2, ldots, i_m]$ for $w$: write $w = r_{i_1}r_{i_2} dotsb r_{i_m}$ with $m$ minimal. Then $partial_w = partial_{i_1}partial_{i_2} dotsb partial_{i_m}.$



In your example, with permutations written in second-line notation, $w_0^3 = 321$ and $pi = 213$, hence $pi^{-1}w_0^3 = 312 = r_2r_1$. Therefore $partial_{pi^{-1}w_0^3} = partial_2partial_1$.







share|cite|improve this answer














share|cite|improve this answer



share|cite|improve this answer








edited May 4 '16 at 16:04

























answered May 4 '16 at 14:02









Catalin ZaraCatalin Zara

3,752514




3,752514












  • $begingroup$
    To Ayush: If my answer is not clear enough for you to accept it, please let me know what else should I add. Thank you.
    $endgroup$
    – Catalin Zara
    Nov 30 '16 at 21:06


















  • $begingroup$
    To Ayush: If my answer is not clear enough for you to accept it, please let me know what else should I add. Thank you.
    $endgroup$
    – Catalin Zara
    Nov 30 '16 at 21:06
















$begingroup$
To Ayush: If my answer is not clear enough for you to accept it, please let me know what else should I add. Thank you.
$endgroup$
– Catalin Zara
Nov 30 '16 at 21:06




$begingroup$
To Ayush: If my answer is not clear enough for you to accept it, please let me know what else should I add. Thank you.
$endgroup$
– Catalin Zara
Nov 30 '16 at 21:06


















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